# Direct sum of injective modules is injective

By the Bass-Papp Theorem, for a unital ring $R$, any direct sum of injective left $R$-modules is injective if and only if $R$ is left Noetherian. I would like to restrict my consideration to an arbitrary abelian subcategory $\mathcal{C}$ of the category $R\text{-mod}$ of unitary left $R$-modules.

We say that an abelian subcategory $\mathcal{C}$ of $R\text{-mod}$ is injectively closed if it satisfies the property that, for an arbitrary family $\left(I_\alpha\right)_{\alpha\in A}$ of injective objects in $\mathcal{C}$ such that $I:=\bigoplus\limits_{\alpha \in A}\,I_\alpha$ is an object in $\mathcal{C}$, $I$ is an injective object in $\mathcal{C}$. Is it true that if $R$ is left Noetherian, then any abelian subcategory of $R\text{-mod}$ is injectively closed? If not, can you please provide a counterexample? Is there a sufficient condition for $\mathcal{C}$ to be injectively closed? References are greatly appreciated.

For a nontrivial example, let $\mathfrak{g}$ be a finite-dimensional semisimple Lie algebra over an algebraically closed field of characteristic $0$ with a triangular decomposition $$\mathfrak{g}=\mathfrak{n}^-\oplus \mathfrak{h}\oplus \mathfrak{n}^+\,.$$ Denote by $\bar{\mathcal{O}}$ the full subcategory of the category of $\mathfrak{U}(\mathfrak{g})$-modules (where $\mathfrak{U}(\mathfrak{g})$ is the enveloping algebra of $\mathfrak{g}$) consisting of $\mathfrak{U}(\mathfrak{g})$-modules $M$ with the following properties:

1. $M$ is a weight module with respect to the Cartan subalgebra $\mathfrak{h}$,

2. each weight space of $M$ is finite dimensional, and

3. $M$ is locally $\mathfrak{n}^+$-finite (that is, $\mathfrak{U}\left(\mathfrak{n}^+\right)\cdot v$ is a finite-dimensional vector subspace of $M$ for any $v\in M$).

Then, $\bar{\mathcal{O}}$ is injectively closed. (In this example, note that $\mathfrak{U}(\mathfrak{g})$ is both left and right Noetherian.)

P.S. I: The Bass-Papp Theorem can be found, for example, in Theorem 3.39 on Page 123 of An Introduction to Homological Algebra by Joseph Rotman.

P.S. II: I posted this question here as well, but didn't receive any answer. I figured that MathOverflow users may be able to help. I don't know how to move my Math.StackExchange post to MathOverflow.

• When you asked this on math.stackexchange, you clarified in response to a comment of Eric Wofsey that you didn't assume the subcategory to be full, and he pointed out that every module category is a (not usually full) abelian subcategory of the category of abelian groups. Doesn't that answer your main question? May 17, 2017 at 8:40
• I'm not sure how it answers my question. It does reduce everything into studying abelian subcategories $\mathcal{A}$ of the category $\mathbf{Ab}$ of abelian groups. Maybe, I'm missing something trivial, but I am not quite sure which subcategories $\mathcal{A}$ are injectively closed. I'm looking for some nontrivial sufficient condition. May 17, 2017 at 10:10
• Take $R=\mathbb{Z}$, so $\mathbf{Ab}=R\text{-mod}$, and take the module category of some non-Noetherian ring as an abelian subcategory of $R\text{-mod}$. May 17, 2017 at 10:14

EDIT Thanks to Jeremy Rickard for several corrective insights in the comments!

In the direction of positive conditions, I'm not sure whether the following conditions are reasonable for your purposes:

Proposition: Let $\mathcal{C}$ be an abelian category which is

• locally finitely presentable (i.e. it has a generator of finitely presentable objects and is complete, or equivalently cocomplete) and

• "strongly Noetherian" in the sense that any subobject of a finitely-presentable object is finitely-presentable, and similarly for higher presentability degrees.

Then $\mathcal{C}$ is injectively closed.

Proof sketch: One shows that in such a category, an object is injective if and only if it lifts against monomorphisms between finitely-presentable objects. This uses the "strong Noetherian" property. This implies, in conjunction with the Noetherianity property again, that injective objects are closed under filtered colimits; since they are also closed under finite direct sums, they are thus closed under arbitrary direct sums.

Corollary: Let $R$ be a "strongly Noetherian ring" in the sense that $R$-$\mathrm{Mod}$ is "strongly Noetherian", and suppose that $\mathcal{C}$ is a full abelian subcategory of $R$-$\mathrm{Mod}$ which is

• closed under kernels and colimits in $R$-$\mathrm{Mod}$, and

• generated under colimits by finitely-presentable $R$-modules.

Then $\mathcal{C}$ is injectively closed.

Proof sketch: Check that $\mathcal{C}$ satisfies the hypotheses of the proposition.

Notes:

1. As Jeremy Rickard points out, for countable $R$, $R$-$\mathrm{Mod}$ is "strongly Noetherian" iff it is Noetherian. So it seems that it is not that much stronger a condition than simply being Noetherian.

2. In Jeremy Rickard's example, where $\mathcal{C}$ is the subcategory of $R$-$\mathrm{Mod}$ given by the image of $S$-$\mathrm{Mod}$, it is the case that $\mathcal{C}$ is closed under kernels and colimits, and is itself locally finitely presentable. But it doesn't satisfy the hypotheses of the corollary because $\mathcal{C}$ is not generated under colimits by modules which are finitely-presentable as $R$-modules. Another way to say this is that the embedding $S$-$\mathrm{Mod} \to R$-$\mathrm{Mod}$ doesn't preserve finite presentability: the module $S$ itself is finitely-presentable as an $S$-module but not as an $R$-module.

• Are you sure that being closed under limits and filtered colimits is sufficient for a full abelian subcategory of the module category of a "strongly Noetherian" ring to satisfy your conditions? You don't need some stronger condition on generators of the subcategory? In the counterexample in my answer, $R$ is a finite-dimensional algebra, and so presumably "strongly Noetherian", and my subcategory is closed under all limits and colimits. May 20, 2017 at 10:14
• Yes, there's definitely an issue -- the notion of "strongly Noetherian" won't get passed from $R$-$\mathrm{Mod}$ to $\mathcal{C}$ unless the notion of finite presentability agrees. So maybe require that $\mathcal{C}$ be generated under filtered colimits by finitely-presentable $R$-modules, and to be safe also assume that $\mathcal{C}$ is closed under cokernels, hence all colimits. But your example has all these properties, so something else must be wrong. Perhaps "strong Noetherianity" is too strong... I bet there are countably-generated abelian groups which are not countably-presentable! May 20, 2017 at 11:54
• If $R$ is a ring with cardinality less than $\kappa$ (a regular cardinal), isn't $\kappa$-presentable for an $R$-module equivalent to "cardinality less than $\kappa$"? And so, for a countable ring $R$, "strongly Noetherian" $\Leftrightarrow$ Noetherian? May 29, 2017 at 17:59
• Oh.... of course you're right! I was thinking initially that something along these lines should hold, but then fell into the trap of thinking that for $R$ countable, a submodule of a countable $R$-module could have presentability rank the continuum, which is nonsense -- it's only the number of subobjects that might be continuum-sized. I see now -- the thing about your example is that the image of $S$-Mod in $R$-Mod is not generated by modules which are finitely-presentable in $S$-Mod. At least, the most obvious generator, $S$ itself, is not finitely-presentable as an $R$-module. May 29, 2017 at 18:55
• There is also a converse statement of sorts in Stenstrom's Rings of Quotients, V. Prop 4.3 : Assuming that $\mathcal{C}$ is locally finitely generated and that $\mathcal{C}$ has enough injectives, the injectives are closed under arbitrary coproducts if and only if $\mathcal{C}$ is locally Noetherian (which I think should be your "Noetherian" after possibly some matching of definitions). Mar 23, 2020 at 14:02

Here's an example of a full exact embedding of the module category of a non-Noetherian ring $S$ into that of a Noetherian ring $R$, preserving all direct sums and direct products. So this gives an example of an abelian subcategory of $R\text{-mod}$ that is not injectively closed, but satisfies many additional "niceness" properties.

Let $k$ be a field, and $R$ the path algebra over $k$ of the quiver with two vertices and three arrows from the first vertex to the second. So an $R$-module is given by the data $(U,V,\alpha,\beta,\gamma)$, where $U$ and $V$ are vector spaces, and $\alpha$, $\beta$ and $\gamma$ are linear maps $U\to V$. Then $R$ is a finite dimensional algebra, and so certainly Noetherian.

Let $S=k\langle x,y\rangle$, the free algebra on two (non-commuting) generators. Then $S$ is not left Noetherian.

The module category of $S$ embeds fully into the module category of $R$ by sending an $S$-module $M$ to the $R$-module given by the data $(M,M,x,y,1_M)$